# Simple Diode circuit

#### prep

Joined Jan 8, 2006
2
I found the answer to this circuit asking Current and Voltage accross the Diode by graphing. A bit rusty in doing numbers , any help appreciated...

the question is attached

Basically need to solve the following eq:
x + 10^-12 * exp( (6.189x10^-18)x ) = 1+10^-12
where x=Vd

already plug in to Matlab and solved it, but wonder how to do by hand...
sorry if this is too trivial...been a while

[attachmentid=1093]

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by prep@Jan 9 2006, 05:34 AM
I found the answer to this circuit asking Current and Voltage accross the Diode by graphing. A bit rusty in doing numbers , any help appreciated...

the question is attached

Basically need to solve the following eq:
x + 10^-12 * exp( (6.189x10^-18)x ) = 1+10^-12
where x=Vd

already plug in to Matlab and solved it, but wonder how to do by hand...
sorry if this is too trivial...been a while

[attachmentid=1093]
[post=12964]Quoted post[/post]​
It is not straigthforward algorithmic solution. The hint is to use iterative/numerical technique. This can be quite laborious.

The approach is to first assume a possible working value of Vd, between 0.6 to 0.7V is pretty reasonable. Solving the equation for current (I) would give a discrete value which can be checked against the total voltage in the loop (1V = I*R + Vd). This wouldn't give a correct value at first, but instead show you how far the assumption is wrong and in what direction.

This is where the iterative process and most of the fun begin. The assumed Vd value has to be corrected and the calculation is performed again. You could see in what direction the results move (smaller or larger error) and by how far. These are used to perform the calculation again and again, each time with improved Vd assumption until you reach an acceptable result (below certain percentage of Vd and I changes from one calculation to the next or oscillating between + and - errors within certain percentage of Vd changes).

I would suggest the easiest approach is to find a low and high Vd that gives positive and negative errors and then halving the value for each iteration.

Example:

1. Assume Vd = 0.4V, calculate I.
2. Assume Vd = 0.8V, calculate I.
3. If assumption (1) under-estimate the Vd and assumption (2) over-estimate the Vd, then the next Vd assumption could be made by halving the span, say Vd=0.6, calculate the I.
4. If assumption (3) under-estimate the Vd, then halving again between Vd=0.6 and Vd=0.8, would give a new assumption of Vd=0.7.
5. Repeat until the desired accuracy is reached.